# I Inverse results in special relativity

#### Kairos

The mass of an object moving at speed v increases such that $$\frac{m'}{m}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

and its apparent frequency decreases such that $$\frac{\nu'}{\nu}=\sqrt{1-\frac{v^{2}}{c^{2}}}$$

so $$\frac{\nu'}{\nu}=\frac{m}{m'}$$

but equating the energies $$h\nu= mc^{2}$$

gives the inverse relationship $$\frac{\nu'}{\nu}=\frac{m'}{m}$$
Has this question already been addressed? Otherwise, thank you in advance if someone can point out my mistake.

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#### weirdoguy

The mass of an object moving at speed v increases
It does not:

• vanhees71

#### DrStupid

The mass of an object moving at speed v increases such that $$\frac{m'}{m}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$
In order to avoid answers like #2 you better repeat your question with energy instead of mass:

$$\frac{E}{E_0}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

$$h\nu= E$$

When you do that you might already see what your problem actually results from.

• vanhees71

#### Kairos

In order to avoid answers like #2 you better repeat your question with energy instead of mass:

$$\frac{E}{E_0}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$
Thanks. Yes, I misspoke; but I still don't understand. This ratio of energies corresponds that of wavelengths, not of frequencies

$$\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\lambda}{\lambda_0}=\frac{\nu_0}{\nu}$$

does it not?

#### Mister T

Gold Member
The mass of an object moving at speed v
Why do you say such an object has a frequency $\nu$? Is it executing simple harmonic motion or something like that? If so, it doesn't have a constant speed.

#### Kairos

Why do you say such an object has a frequency $\nu$? Is it executing simple harmonic motion or something like that? If so, it doesn't have a constant speed.
OK I suppose I see what you mean: the above relationship between energies does not hold for a harmonic resonator in a rocket at speed v?

#### Orodruin

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To the extent that the de Broglie relation can be generalised to 4-momentum and 4-frequency, your relation between the frequencies is wrong.

#### JD_PM

its apparent frequency decreases
Note that when I use apparent I mean I am measuring the whatever quantity associated to the rocket wrt $S$. (stationary frame).

I assume that we are dealing with, let's say, a rocket with a flashing light on it getting further and further away from a stationary frame (if it were approaching it instead, the apparent frequency $\nu$ would increase; see Dopler effect: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect).

The apparent wavelength will increase; let me explain why.

A time $t$ elapses between flashes, so during this time the light will have traveled a distance $ct$. But, of course, during that time the spaceship has also traveled a distance: $ut$. Then the apparent wavelength will be:

$$\lambda = (c + u)t$$

Note that $t$ is the time as measured by an observer in $S$ (it's the 'apparent' time). As moving clocks run slow:

$$t = \gamma \bar t$$

Where $\bar t$ is the time as measured by an observer in $\bar S$.

Then the apparent frequency is:

$$\nu = \frac{c}{\lambda} = \frac{c}{(c + u)t} = \frac{c(\sqrt{1 - u^2/v^2})}{(c + u)\bar t} = \frac{\sqrt{c - u}}{\sqrt{c + u}}\bar \nu$$

Where $\bar \nu$ is the rest frequency (I have used $\bar t = \frac{1}{\bar \nu}$).

Actually professor Michel van Biezen works out the same exercise but with an approaching spaceship:
Hope at this point it is more clear why the frequency decreases wrt $S$ frame.

#### Mister T

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OK I suppose I see what you mean: the above relationship between energies does not hold for a harmonic resonator in a rocket at speed v?
That is not my point. My point is that you have an object of mass $m$ moving with speed $v$. Such an object will indeed have a rest energy of $mc^2$. But you lose me when you say its energy is equal to $h\nu$. You don't tell us what $\nu$ is. It's a frequency, but the frequency of what?

$h\nu$ is the energy of a photon, but a photon has zero mass, thus $mc^2=0$.

#### Mister T

Gold Member
I assume that we are dealing with, let's say, a rocket with a flashing light on it getting further and further away
That can't be the case. The mass $m$ of the rocket is not in any way related to the frequency $\nu$ of the flashing light. In other words, there is no relation between $m$ and $\nu$, therefore expressions like $mc^2=h\nu$ can't be valid and are founded on nonsense.

#### JD_PM

It's true that I just focused on the frequency. Maybe a proton moving with a relativistic speed is a good example here.

But let's wait for clarification.

#### Orodruin

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That can't be the case. The mass $m$ of the rocket is not in any way related to the frequency $\nu$ of the flashing light. In other words, there is no relation between $m$ and $\nu$, therefore expressions like $mc^2=h\nu$ can't be valid and are founded on nonsense.
Are you calling the de Broglie relation nonsense? I mean, seriously outdated, fine, but nonsense seems to be stretching it a bit far.

• PeroK

#### pervect

Staff Emeritus
$E = h \, \nu$ applies to a photon, but the frequency shift for a photon is given by the relativistic doppler formula

$$\frac{f_s}{f_r} = \sqrt{\frac{1+\beta}{1-\beta}}$$

This is important to understand for the correct expression of how the energy and momentum of a photon transforms relativistically.

For non-photons, if you want to consider relativistic effects, you need to use a relativistic form of quantum mechanics. For instance, you could use the Klein-Gordon equations for a particle with no spin, or the Dirac equations for a particle with spin 1/2.

Wiki has <<this link>> on relativistic wave equations. I'm not really all that familiar with the topic, unfortunately, and I can't say how good it is. You could try the quantum forum if you need a better answer.

Apparently, though, one can still use the Schrodinger wave equation

$$i\,h\frac{\partial}{\partial t} \, \Psi = H \, \Psi$$

but the relativisitc Hamiltonian operator, H, is not the non-relativistic one involving $\frac{\hbar^2}{2m}$. Basically, this form of the hamiltonian comes from $p^2/2m$, where p is the momentum, and you can see that it's a classical expression, not a relativistic one.

#### Kairos

That is not my point. My point is that you have an object of mass $m$ moving with speed $v$. Such an object will indeed have a rest energy of $mc^2$. But you lose me when you say its energy is equal to $h\nu$. You don't tell us what $\nu$ is. It's a frequency, but the frequency of what?

$h\nu$ is the energy of a photon, but a photon has zero mass, thus $mc^2=0$.
Understood. The form of Energy
$h\nu$ is not generalisable to objects with a mass through de Broglie wavelength?

#### Orodruin

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Understood. The form of Energy
$h\nu$ is not generalisable to objects with a mass through de Broglie wavelength?
As far as the de Broglie relation is applicable (with all the caveats of being outdated as per the above remarks), yes it is. The relativistic variant of the de Broglie relation is $h N = P = mV$, where $N$ is the 4-frequency, $P$ the 4-momentum, $m$ the invariant mass, and $V$ the 4-velocity of the object. Note that I am using units where $c = 1$.

#### Orodruin

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Thanks. Yes, I misspoke; but I still don't understand. This ratio of energies corresponds that of wavelengths, not of frequencies

$$\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\lambda}{\lambda_0}=\frac{\nu_0}{\nu}$$

does it not?
It is unclear to me why you think that
$$\frac{\lambda}{\lambda_0} = \gamma(v).$$
This is not the case for a matter wave, you cannot apply length contraction on wavelengths.

#### Orodruin

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Why would you? There is not a single thing about a matter wave that lines up with length contraction for the wavelength. The phase velocity is > c.

#### JD_PM

Why would you?
I asked because while working out post #8 I thought 'why not $\lambda = \gamma (c + u)t$?; $(c + u)t$ has dimensions of length, and we know that moving objects get shortened along the direction of motion...'.

#### Orodruin

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I asked because while working out post #8 I thought 'why not $\lambda = \gamma (c + u)t$?; $(c + u)t$ has dimensions of length, and we know that moving objects get shortened along the direction of motion...'.
You really should stop thinking in terms of length contraction and time dilation if you cannot easily identify the cases where they are applicable and those where they are not. Matter waves have no ”rest frame”. Their wavelength when in the rest frame of the corresponding object is infinite.

• • SiennaTheGr8 and JD_PM

#### Mister T

Gold Member
Are you calling the de Broglie relation nonsense?
No, of course not. I didn't realize that $\nu$ was a frequency associated with the de Broglie wavelength.

If that's the case, then the derivation presented in the OP is flawed because the energy $E$ of an object of mass $m$ moving with speed $v$ is not equal to $h\nu$.

• Kairos

#### SiennaTheGr8

You really should stop thinking in terms of length contraction and time dilation if you cannot easily identify the cases where they are applicable and those where they are not. Matter waves have no ”rest frame”. Their wavelength when in the rest frame of the corresponding object is infinite.
Indeed. I've seen experts casually make this mistake in books, chalking something up to length contraction when it doesn't apply. I can think of two examples off the top of my head, actually. If unsure, identify events and use the Lorentz transformation!

#### Kairos

Going back to my original problem, by combining posts 3 and 4,

$\frac{E}{E_0}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\lambda}{\lambda_0}=\frac{\nu_0}{\nu}$

when the energies are rather supposed to be proportional to the frequencies through $E= h\nu$
Does it mean that the first identity holds for a mass but is inverted for a photon?

#### Orodruin

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Going back to my original problem, by combining posts 3 and 4,

$\frac{E}{E_0}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\lambda}{\lambda_0}=\frac{\nu_0}{\nu}$

when the energies are rather supposed to be proportional to the frequencies through $E= h\nu$
Does it mean that the first identity holds for a mass but is inverted for a photon?
No. Again, your relationship between the wavelengths is wrong. In addition, $\lambda/\lambda_0 = \nu_0/\nu$ only holds if the speed of the wave is independent of the reference frame, which is also not the case.

#### pervect

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Why would you? There is not a single thing about a matter wave that lines up with length contraction for the wavelength. The phase velocity is > c.
I would argue that if one is using a relativistic formulation of quantum mechanics (and I would guess the Original Poster is most likely not doing this), then the wave function of a single particle should transform via the Lorentz transform.

As an aside, I'm not quite sure of the details of how a wavefunction of a multi-particle system would transform. But let's stick to single particle wavefunctions for the time being.

I would expect that that Lorentz contraction would be part of (but not the complete description of) how the wavefunction transforms.

It's possible that some confusion could arise from omitting the other features of the Lorentz transform other than length contraction (time dilation and the relativity of simultaneity, to be specific). I don't think that's the main problem here, though.

I suppose a more specific problem could be useful, one might imagine asking what the allowed energy states of a particle in a box were, and perversely insist on working the problem out in a frame of reference where the box was moving at relativistic velocities, rather than in the rest frame of the box.

The box would length contract in such a scenario, and so would the wavefunction of the particle, which would be bounded if one makes the usual assumption of the box being an infinitely deep potential well.

Of course it'd be wrong to just consider length contraction of the box, and ignore other aspects arising from the Lorentz transform, but I don't think that's necessarily the issue here.

The usual textbook treatment of QM, not being a relativistic formulation, would probably not give covariant answers for the problem of a particle in a relativistic box.

"Inverse results in special relativity"

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